已知正项等差数列an的公差不为0,a2,a5,a14是等比数列bn的前三项

a1=25、a11=25＋10d、a13=25＋12d则：a11²=(a1)×(a13)(25＋10d)²=25×(25＋12d)得：d=－2则：a(n)=－2n＋27数列a1、a

1.因为等差数列AN的公差d不等于0,a1=2,s9=36,所以36=9*2+1/2*9*8d所以d=1/2所以a3=3,a9=6,由a3,a9,am成等比数列则a9的平方=a3*am,的am=12又

已知﹛an﹜是首项为-16,公差不为0的等差数列,其前n项和为Sn,且a1,a5,a4成等比数列,求﹛an﹜的公差d.设a1=-16a5=-16+4da4=-16+3d所以,a5²=a1*a

设公差为d则a3=a1+2d=1+2da9=a1+8d=1+8d因为a1,a3,a9成等比数列所以a3²=a1*a9=a9∴(1+2d)²=1+8d∴d=0或者d=1又∵d≠0,∴

an=a1+(n-1)d=2+(n-1)da2=2+da4=2+3da8=2+7da2,a4,a8成等比数列,即a4/a2=a8/a4a4*a4=a2*a84+12d+9d^2=4+16d+7d^22

（1）根据题意,设公差为d则a3=a1+2d=2d+1a9=a1+8d=8d+1有(2d+1)^2=8d+1d=1故通项：an=n（2）根据题意,设公比为q则b2=qb3=q^2有q-0.5q^2=0

（Ⅰ）由题知a\x0523=a1a7,设等差数列{an}的公差为d,则(a1+2d)2=a1（a1+6d）,a1d=2d2,∵d≠0∴a1=2d．…（1分）又∵a2=3,∴a1+d=3a1=2,d=1

a9=a5+4da15=a5+10d(a5+4d)²=a5(a5+10d)8da5+16d²=10da516d²-2da5=02d(8d-a5)=0d=a5/8所以a9=

1)设公差为d已知(a4)^2=a2*a5则(a1+3d)^2=(a1+d)(a1+4d)a1*d+5d^2=05d=-a1=10d=2故通项公式an=-10+2(n-1)=2n-122)bn=a^[

a3=b3a1+2d=b1*q^2=a1*q^2a1+2d=a1*q^2.1a7=b5a1+6d=b1*q^4=a1*q^4a1+6d=a1*q^4.21式×3-2式2a1=3a1*q^2-a1q^4

设公差为d,则d≠0a1,a3,a9成等比数列,则a3²=a1·a9(a1+2d)²=a1(a1+8d)a1=1代入,整理,得d²-a1d=0d(d-a1)=0d≠0,因

由a6=23＋5d＞0和a7=23＋6d＜0,得公差d=－4由a6＞0,a7＜0,∴S6最大,S6=8.由a1=23,d=－4,则Sn=1/2n(50－4n),设Sn＞0,得n＜12.5,整数n的最大

因为{An}是等差数列,所以A2+A8=A4+A6=10,A4*A6=24,所以可将A4、A6看作方程x^2-24x+10=0的两个根,因为d

先做个mark,回头再做给你看.----------------------------------------将{an}分拆成{bt}、{ct}数列排列如下：{bt}:a1,a3,a5,a7,a9,

S1/a1=1S2/a2-S1/a1=(2+d)/(1+d)-1=d/(1+d)S3/a3-S1/a1==(3+3d)/(1+2d)-1=(2+d)/(1+2d)2*d/(1+d)=(2+d)/(1+

a2=a1+da4=a1+3da6=a1+5da2,a4－2,a6成等【比】数列(a1+3d-2)^2=(a1+d)(a1+5d)(3d-1)^2=(1+d)(1+5d)9d^2-6d+1=5d^2+

设该等差数列是首项为a1,公差为dS3=3a1+3(3-1)*d/2=3a1+3dS2=2a1+2(2-1)*d/2=2a1+dS4=4a1+4(4-1)*d/2=4a1+6d又:S3²=9

设an=a1+(n-1)d则a2=a1+da3=a1+2da4=a1+3da7=a1+6d因为等差数列{an}的前四项和为10所以,a1+a2+a3+a4=10即4a1+6d=10.①又因a2,a3,

由a6=23＋5d＞0和a7=23＋6d＜0,得公差d=－4由a6＞0,a7＜0,∴S6最大,S6=8.由a1=23,d=－4,则Sn=1/2n(50－4n),设Sn＞0,得n＜12.5,整数n的最大